Question

The probability that a randomly selected education major received a starting salary offer greater than $52,350 is . The probability that a randomly selected education major received a starting salary offer between $45,000 and $52,350 is . (Hint: The standard normal distribution is perfectly symmetrical about the mean, the area under the curve to the left (and right) of the mean is 0.5. Therefore, the area under the curve between the mean and a z-score is computed by subtracting the area to the left (or right) of the z-score from 0.5.) What percentage of education majors received a starting offer between $38,500 and $45,000? 6.68%

Answer 1

Answer:

• 1.  0.0808

• 2. 0.5371

• 3. 58.2%

Explanation:

The first part of the question, with the input data, is missing.

This is missing part:

• Assume that X, the starting salary offer for education majors, is normally distributed with a mean of $46,292 and a standard deviation of $4,320

Solution

Question 1: The probability that a randomly selected education major received a starting salary offer greater than $52,350 is ________

Find the Z-score.

The Z-score is the standardized value of the random variable and represents the number of standard deviations the value of the random variable is away from the mean:

       $Z-score=\dfrac{X-\mu}{\sigma}$

      $Z-score=\dfrac{\ 52,350-\ 46,292}{\ 4,320}=1.40$

      $P(X>\ 52,350)=P(Z>1.40)=P(Z<-1.40)$

The standard normal distribution tables give the cumulative probabilities as the cumulative area under the bell-shaped curve. P(Z>1.40) is the area under the curve that is to the right of Z = 1.40 or, what is the same, P(Z<-1.40) is the area to the left of Z = - 1.40.

Using the former, the table indicates P(Z>1.40) = 0.0808, which is 8.08%

Question 2: The probability that a randomly selected education major received a starting salary offer between $45,000 and $52,350 is _______.

Now you  must find the area under the curve between the two Z-scores.

    $Z(X= 45,000)=\dfrac{\ 45,000-\ 46,292}{\ 4,320}=-0.30$

   $Z(X= 52,350)=\dfrac{\ 52,350-\ 46,292}{\ 4,320}=1.40$

Then, you must find the area between Z = -0.30 and Z = 1.40

That is P(Z<1.40) - P(Z < - 0.30) = P(Z > - 1.40) - P(Z < - 0.30)

                                                    = 1 - P(Z< -1.40) - P( Z < -0.30)

Now, you can work with the area under the curve to the left of Z = - 1.40 and to the left of Z = - 0.30.

From the corresponding table, that is: 1 - 0.0808 - 0.3821 = 0.5371

Question 3. What percentage of education majors received a starting offer between $38,500 and $45,000?

For X = $38,500:

      $Z=\dfrac{\ 38,500-\ 46,292}{\ 4,320}=-1.80$

For X = $45,000

     Z = -0.3 (calculated above)

Then, you must find the area under the curve to the right of Z = - 1.80 and to the left of Z = - 0.3

• P (Z > - 1.80) - P (Z < - 0.3)

• 1 - P (Z < - 1.80) - P (Z < - 0.3)

• 1 - 0.0359 - 0.3821 = 0.582 = 58.2%